Sachin throws a ball vertically upwards. The ball returns to sachin after 6 sec. Can you find :1) the velocity with which sachin throws the ball2) the maximum height it reaches3) its position after 4 sec Share with your friends Share 7 Vara answered this (1)Time of flight, T=2ugVelocity of projection:u=gT2 =9.8×62 =29.4 m/s(2)Maximum height:H=u22g =29.422×9.8 =44.1 m(3)Position after 4 sec:S=ut-12gt2 =29.4×4-12×9.8×42 =39.2 m from the ground 3 View Full Answer B Asha answered this a] - 60ms-1b] 180m c] 100 m i think this is the answer -1 Ira Dadhe answered this Please explain the way u took out the solutions..... 4